Question

1.99 m/s^2 for the first 60.0 m, then decelerates at -0.266 m/s^2 for the final 40.0 m. How much time did the race take?

Answer 1

Time required to complete 60m:-

`\\ \rm\longmapsto s=1/2at^2`

`\\ \rm\longmapsto 60=1.99/2t^2`

`\\ \rm\longmapsto 120=1.99t^2`

`\\ \rm\longmapsto t^2=60.3`

`\\ \rm\longmapsto t\approx 7.7s`

Time taken to complete last 40m

`\\ \rm\longmapsto 80=-0.266t^2`

`\\ \rm\longmapsto t^2=300.75`

`\\ \rm\longmapsto t\approx 17.3s`

Total time:-

• 7.7+17.3=25s(Approx)

Answer 2

Step-by-step explanation:

Two kinematics equation

v = u + at

and

s = ut + 1/2(at^2)

where u is initial velocity, v is final velocity, a is acceleration, t is time and s is displacement.

For the first 60m, initial velocity is 0 and acceleration is 1.99.

60 = 0 + 1/2(1.99)t^2

t^2 = 120/1.99

t = 7.765s

Velocity at 60m = 0 + 1.99(7.765) = 15.453 m/s

For the final 40m, initial velocity = 15.453, acceleration is -0.266.

40 = 15.453t' + 1/2(-0.266)t'^2

0.133t'^2 - 15.453t' + 40 = 0

Using the quadratic formula, t' = 2.649s or 113.54s

The solution of 113.54s represents someone running pass the finish line and continuing the deceleration until one runs backwards to the finish line again. So that solution should be rejected.

The total time taken = 7.765 + 2.649

= 10.414s